\beginsection 2.1

Show that $\sqrt3$, $\sqrt5$, $\sqrt7$, $\sqrt{24}$, and
$\sqrt{31}$ are not rational numbers.

\medskip
The solution to $x^2-3=0$ is $\sqrt3$.
By the Rational Zeros Theorem, if $\sqrt3=p/q$ then
$q$ divides $a_1=1$ and $p$ divides $a_0=3$.
Since $q$ must be $\pm1$, the only possible values for
$p/q$ are $\pm1$ and $\pm3$, none of which are
solutions to $x^2-3=0$.
Therefore $\sqrt3$ must be irrational.

$\sqrt5$: $p/q=\pm1,\pm5$, none of which are
solutions to $x^2-5=0$.

$\sqrt7$: $p/q=\pm1,\pm7$, none of which are
solutions to $x^2-7=0$.

$\sqrt24$: $p/q=\pm1,\pm2,\pm3,\pm4,\pm6,\pm8,\pm12$,
none of which are solutions to $x^2-24=0$.

$\sqrt31$: $p/q=\pm1,\pm31$, none of which are
solutions to $x^2-31=0$.
